Does the Laplace operator have a fundamental solution on the surface of a cylinder in $\mathbb{R}^3$?
Intuitively, I can visualize a function that diverges to $\infty$ at a point, decreases to a saddle point at the antipodal point, and decreases to $-\infty$ in both directions along the cylinder axis. But I don't even know how to go about proving whether a fundamental solution exists -- since the cylinder is not simply connected, I can't conformally map it to the disk, for instance.
My question is a special case of a previous unanswered question, for which I'm also going to add a bounty.
The cylinder (of unit radius, for simplicity) is obtained from the infinite strip $S = \{z\in \mathbb C: |\operatorname{Im}z|<\pi \}$ by identifying its sides. What you need is Neumann Green's function for $S$ (i.e., with zero normal derivative on $\partial S$), because it will remain harmonic after gluing the sides. And to find this function, you can use conformal maps: under a conformal map, normal derivative is multiplied by a positive factor.
On the right half-plane $P$, the Neumann Green's function with pole at $1$ is $$u(z)= -\frac{1}{2\pi} \log|(z-1)(z+1)| = -\frac{1}{2\pi} \log| z^2-1|$$ (compare this to $ -\frac{1}{2\pi} \log\frac{|z-1|}{|z+1|}$ for the Dirichlet condition). The exponential map $z\mapsto \exp(z/2)$ sends $S$ onto $P$. Therefore, $$-\frac{1}{2\pi} \log|e^z-1|$$ does the job on $S$ (with pole at $0$). However, the lack of left-right symmetry is annoying. It's better to average this function with its reflection $-\frac{1}{2\pi} \log|e^{-z}-1|$, thus arriving at $$g(z,0) = -\frac{1}{4\pi} \log|(e^z-1)(e^{-z}-1)| = -\frac{1}{4\pi} \log|2 - 2 \cosh z| $$
This looks like you expected, if you imagine $y=\pm \pi$ being glued together:
By the way: as the radius of cylinder shrinks to $0$ (equivalently, we take a large-scale view of the above function), we approach one-dimensional Green's function, which can be written as a multiple of $-|x|$ (in the original, non-symmetric form, we'd have a multiple of $\min(0,-x)$ instead).